# The Borel class of a subset of $\mathbb Z^\omega$

Define $$F(t)=\ln(t+1)$$ for $$t\geq 0$$.

For each sequence of integers $$s=s_0s_1s_2...\in \mathbb Z^\omega$$ define $$t^*_{ s}=\sup_{n\geq 0}F^{n}(|s_n|)$$ where $$F^{n}$$ is the $$n$$-fold composition of $$F$$.

Let $$\sigma$$ be the shift mapping on $$\mathbb Z^\omega$$; so $$\sigma(s_0s_1s_2...)=s_1s_2s_3...$$., and let $$\sigma^k$$ be the $$k$$-fold composition of $$\sigma$$.

Is the set $$\mathbb S:=\{s\in \mathbb Z^\omega:t^*_{\sigma^k(s)}\to\infty \text{ as }k\to\infty\}$$ an $$F_{\sigma\delta}$$-subset of $$\mathbb Z ^\omega$$? Assume $$\mathbb Z$$ is given the discrete topology and $$\mathbb Z ^\omega$$ has the product topology.

A positive answer to this question would imply that a certain set in complex dynamics is an Erdős space factor. See this paper, Remark 5.3 in particular, for more about this problem. Essentially a space $$E$$ is an Erdős space factor if $$E\times \mathfrak E\simeq \mathfrak E$$ where $$\mathfrak E$$ is the rational Hilbert space.

EDIT 10/26/20: I proved that the space from complex dynamics (mentioned above) is not an Erdős space factor, answering a question by Dijkstra and van Mill: link to paper. This result implies in particular that the set $$\mathbb S$$ is not $$F_{\sigma\delta}$$.

• It seems that you obtain $F_{\sigma\delta}$ writing down the definition of your set: $\bigcap_{n\in\mathbb N}\bigcup_{k\in\mathbb N}\bigcap_{m\ge k}\{s\in\mathbb Z^\omega:s(m)\ge n\}$. – Taras Banakh Jul 12 '20 at 9:48
• @TarasBanakh Your set is a proper subset of the one I wrote. For instance I believe the sequence $0,1,0,2,0,3,0,4...$ belongs to my set but not yours. – D.S. Lipham Jul 12 '20 at 16:14
• The sequence of $t^*$'s for $0,1,0,2,...$ would be above the sequence $F^{1}(1), 1, F^{1}(2), 2, F^{1}(3),3, …$ which goes to $\infty$ – D.S. Lipham Jul 12 '20 at 16:41
• @TarasBanakh I wonder if $t^*\to\infty$ is the same as saying that a subsequence of s goes to infinity? In this case, my set is probably not $F_{\sigma\delta}$. Assuming these sets are equal, how would you prove that? By the way, I have shown that the original set is both $F_{\sigma\delta\sigma}$ and $G_{\delta\sigma\delta}$. – D.S. Lipham Jul 12 '20 at 16:56
• Definitely this is not the same as being unbounded, as suggested in the penultimate comment, just insert sufficiently long zero runs in the sequence $\mathbb{N}$. – Ville Salo Jul 12 '20 at 20:34

Here's what I had in mind. Consider a $$\Sigma^0_3$$-set $$T = \bigcup_{n \in \mathbb{N}} \bigcap_{m \in \mathbb{N}} \bigcup_{k \in \mathbb{N}} C_{n,m,k}.$$ where each $$C_{n,m,k} \subset 2^\mathbb{N}$$ is clopen. We show that there is a continuous map $$f : 2^{\mathbb{N}} \to \mathbb{Z}^\omega$$ such that $$f^{-1}(S) = T$$ where $$S = \{s \in \mathbb{Z}^\omega \;:\; t^*_{\sigma^k(s)} \rightarrow \infty \text{ as } k \rightarrow \infty \}$$ is the set from the question. This proves that $$S$$ is not $$F_{\sigma \delta}$$, since that would imply all $$\Sigma^0_3$$ sets $$T$$ in $$2^\mathbb{N}$$ are $$\Pi^0_3$$.

You are at the concert. On the stage, there is a conductor and $$\omega$$ many cellists. The conductor is reading a point $$x \in 2^\mathbb{N}$$. Whenever she notices $$x \in C_{n,m,k}$$, the conductor cues the $$n$$th cellist to play the note $$m$$, assuming it hasn't been played before, and $$n$$ has played all the notes before $$m$$. Only one cellist plays at a time, there is a rest when all the events $$x \in C_{n,m,k}$$ visible so far are exhausted, and if $$m$$ cannot be played yet because previous notes have not been played, the conductor makes a note of it and it is played once they have. As you listen to these rising scales, you note that $$x \in T$$ if and only if one of the cellists plays the entire scale $$\mathbb{N}$$.

From this continuously revealed information you will construct the continuous function $$f : 2^\mathbb{N} \to \mathbb{Z}^\omega$$. The construction for $$f(t) = s$$ is thusly. We go through $$\ell = 0, 1, ...$$ and by default we just set $$s_\ell = 100$$ for all $$\ell$$. Whenever the $$n$$th cellist plays, we do as follows:

• if one of the cellists $$n' < n$$ has played between the last time the $$n$$th cellist played (or the beginning of time if $$n$$ hasn't played anything) and the present time, then we set $$s_{\ell} = 100$$. As long as no cellist $$n' < n$$ plays again we ensure that also $$t^*_{\sigma^\ell(s)} = 100$$.
• otherwise (if no cellist $$n' < n$$ has played between), then if the last time $$n$$ played we set $$s_{\ell'} = 100+h$$ then we now set such a high value at $$s_\ell$$ that we have $$\lfloor t^*_{\sigma^{\ell'+1}(s)} \rfloor \geq 100+h+1$$. Namely, set $$s_\ell = \lceil \text{pexp}^{\ell-\ell'-1}(100+h+1) \rceil$$ where $$\text{pexp}(x) = \exp(x) - 1$$. Note that in fact we get precisely $$\lfloor t^*_{\sigma^{\ell'+1}(s)} \rfloor = 100+h+1$$ because of basic properties of $$\text{pexp}$$. It also follows, because $$\log (103 + h) < 100 + h$$ and by induction, that we do not disturb any of the ensured values $$t^*_{\sigma^\ell(s)} = 100$$ for any $$n' \leq n$$: those were ensured before setting the value of $$s_{\ell'}$$.

Now suppose indeed $$t \in T$$, and some cellist plays infinitely many times. Then if the $$n$$th cellist is the first cellist that does, then the first item applies only finitely many times for $$n$$, and after that whenever we set $$s_{\ell} = \text{pexp}^{\ell-\ell'-1}(100+h+1)$$ we actually set $$t^*_{\sigma^{\ell''}(s)} \geq 100+h+1$$ for all $$\ell'' \in [\ell'+1, \ell]$$. So since $$n$$ plays infinitely many times, actually $$t^*_{\sigma^{\ell''}(s)} \rightarrow \infty$$ as $$\ell'' \rightarrow \infty$$.

Suppose then $$t \notin T$$. If the song is finite, obviously $$\lim_\ell t^*_{\sigma^{\ell}(s)} = 100$$. Otherwise, whenever $$n$$ plays for the last time, we have a new ensured value at which $$t^*_{\sigma^{\ell}(s)} = 100$$, thus $$\liminf_\ell t^*_{\sigma^{\ell}(s)} \leq 100$$.

An observation:

• As far as I can tell all we are using is that $$F$$ is monotone, $$F(h + 2) < h$$ for $$h \geq 100$$ and that $$\lfloor F^n(h) \rfloor \rightarrow \infty$$ as $$h \rightarrow \infty$$ for any $$n$$, and the values range over $$[100, \infty) \cap \mathbb{N}$$. And I guess $$100$$ can be replaced by some other number (probably $$1$$ or $$2$$ for your function). Maybe I missed some axioms.

edit

Your set is not $$G_{\delta \sigma}$$ either. Namely, any $$\Pi^0_3$$ set $$T' = \bigcap_{n \in \mathbb{N}} \bigcup_{m \in \mathbb{N}} \bigcap_{k \in \mathbb{N}} D_{n,m,k}$$ clearly continuously reduces to $$S' = \{s \in \mathbb{Z}^\omega \;:\; \lim_i s_i = \infty\}.$$ To see this, for each $$n$$ separately go through $$(m,k)$$ in lexicographical order, advancing to the next $$m$$ when you observe the point is not in $$D_{n,m,k}$$. On step $$\ell$$, output $$n$$ if $$m$$ is updated for $$n$$, otherwise output $$\ell$$. This way you construct $$g(t) \in \mathbb{Z}^\omega$$ for $$t \in 2^\mathbb{N}$$.

Clearly $$t \in T'$$ if and only if $$m$$ is updated for each $$n$$ only finitely many times. If $$m$$ is updated infinitely many times for $$n$$, then the limit of $$g(t)$$ is at most $$n$$, while if $$m$$ is updated finitely many times for each $$n = 0, 1, ..., N$$ then $$g(t)$$ stays above $$N$$ from some point on.

Now it's easy to further reduce to your set, observing that if a sequence satisfies $$|s_{i+1} - s_i| \in \{-1,0,1\}$$ and $$s_i \geq 100$$ for all $$i$$, then $$s \in S' \iff s \in S$$. Just replace all jumps by arithmetic progressions.

By taking the coordinatewise minimum of this process and the above, I suppose we have

Every set of the form $$A \cap B$$ for $$A \in \Sigma^0_3$$ and $$B \in \Pi^0_3$$ continuously reduces to your set.

But I don't know if your set can be written as $$C \cap D$$ for $$C \in F_{\sigma \delta}$$ and $$D \in G_{\delta \sigma}$$.

• Thank you for your answers! I will try to understand soon. – D.S. Lipham Jul 13 '20 at 14:40